An object is thrown vertically from a height of #5 "m"# at #3 "m/s"#. How long will it take for the object to hit the ground?

1 Answer
Mar 4, 2016

Answer:

#0.7 quad "s"#

Explanation:

"Thrown vertically" could mean either upwards or downwards. I am assuming it to be downwards.

Step 1: Understand the problem.

An object is under free fall. It is accelerating at #g = 9.81 quad "m/s"^{2}# downwards. It is given an initial velocity of #3 quad "m/s"# downwards.

Step 2: Identify the relevant equations.

To find displacement, #s#, under constant acceleration, #a#, and initial velocity, #v_0#, after time, #t#, the following equation is applicable:

#s = v_0 t + 1/2 a t^2#

To get the time as a function of the distance traveled, make #t# as the subject of formula. Begin by writing the quadratic equation in standard form.

#t^2 + frac{2 v_0}{a}t - frac{2 s}{a} = 0#

Subsequently, complete the square.

#(t + frac{v_0}{a})^2 - frac{2 s}{a} = (frac{v_0}{a})^2#

#(t + frac{v_0}{a})^2 = frac{v_0^2}{a^2} + frac{2 s}{a}#

#= frac{v_0^2 + 2as}{a^2}#

Reject the negative square root and take the positive one. This is because we only restrict ourselves to #t > 0#.

#t + frac{v_0}{a} = frac{sqrt{v_0^2 + 2as}}{a}#

#t = frac{sqrt{v_0^2 + 2as} - v_0}{a}#

Step 3: Set up the coordinate system.

I let the downwards direction be positive. The ground level is arbitrarily set to be the origin (i.e. #s = 0#). This is up to your preferences.

Step 4: Identiy the parameters.

In this question,

  • #a = g = 9.81 quad "m/s"^2#

  • #v_0 = 3 quad "m/s"#

  • #s = 5 quad "m"#

Step 5: Plug into the equation.

#t = frac{sqrt{v_0^2 + 2as} - v_0}{a}#

#= frac{sqrt{(3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")} - (3 quad "m/s")}{9.81 quad "m/s"^2}#

#= 0.749 quad "s"#

Now earlier, I assume that the object has an initial velocity of #3 quad "m/s"# downwards. The formula also works if the initial velocity is #3 quad "m/s"# upwards. In that case, everything remains the same, except

  • #v_0 = - 3 quad "m/s"#

Plugging in the values again

#t = frac{sqrt{v_0^2 + 2as} - v_0}{a}#

#= frac{sqrt{(-3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")} - (-3 quad "m/s")}{9.81 quad "m/s"^2}#

#= 1.361 quad "s"#